A circle is divided into 6 sectors in such a way that the angles of the sectors are in arithmetic progression. The angle of the largest sector is 4 times the angle of the smallest sector. Given that the radius of the circle is 5 cm, find the perimeter of the smallest sector.
- (a)A circle is divided into 6 sectors in such a way that the angles of the sectors are in arithmetic progression. The angle of the largest sector is 4 times the angle of the smallest sector. Given that the radius of the circle is 5 cm, find the perimeter of the smallest sector.[6]
- (b)(i)the value of the constant k,[3]
- (b)(ii)the sum to infinity of the progression.[2]
Mark scheme
Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.
- [6] (a) a+5d = 4a or (a+4a)/2 × 6; B1 co; Sum of angles = 360° or 2π rads; (2a+5d)/2 × 6 = 360 or (a+4a)/2 × 6 = 360; M1 A1 Correct left-hand side. All correct.; Sim Eqns → a = 24° or 2π/15 rads; A1 Either answer.; Arc length = rθ = 5 * (2π/15); M1 Correct use of arc length with θ in rads.; Perimeter = rθ + 2r = 5 * (2π/15) + 2*5 = 12.094...; A1 co; Perimeter = 12.1.
- [3] (b)(i) k+6 / (2k+3) = k / (k+6); M1 A1 Correct eqn for k.; → k²-9k-36= 0 → k = 12; A1 Co condone inclusion of k = -3. (NB stating a, ar, ar² as f(k) gets M1)
- [2] (b)(ii) r = 18/27 = 2/3, a = 27; M1 A1 Correct formula for S∞ must have -1 < r < 1. co; → S∞ = 27/(1-2/3) = 81.
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →